one sees that $\Omega(A)[0]=2A[0]$.
So, an equation $\Omega(A)=B$
with $B[0]=0$
implies that $A[0]=0$.

Now the operator $\Omega$
acts on series with zero constant term as $\Omega=I+N$
with $I$ identity and $N(A)=(1+t)A(t^2)$ which is
topologically nilpotent. Then
$$
\Omega^{-1}=I-N+N^2-N^3+\ldots
$$
In this case $\Omega(A)=B$ (in case $B[0]=0$
which is your case) has only one solution which is

(infinite sum).
This is easy to program and gives all asymptotic expansions of equations of type
$$
A(t)+(1+t)A(t^2)=B\ ;\ B[0]=0
$$
I tried it for $B(t)=\frac{t}{1-t^2}$ (your question) and $B(t)=sin(t)$.